121616949-math.126 - 112 Chapter 6 Applications of the...

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112 Chapter 6 Applications of the Derivative to maximize. In this problem, the condition can be read off the above drawing: the upper corner of the triangle, whose coordinates are ( h - R, r ), must be on the circle of radius R . That is, ( h - R ) 2 + r 2 = R 2 . We can solve for h in terms of r or for r in terms of h . Either involves taking a square root, but we notice that the volume function contains r 2 , not r by itself, so it is easiest to solve for r 2 directly: r 2 = R 2 - ( h - R ) 2 . Then we substitute the result into πr 2 h/ 3: V ( h ) = π ( R 2 - ( h - R ) 2 ) h/ 3 = - π 3 h 3 + 2 3 πh 2 R We want to maximize V ( h ) when h is between 0 and 2 R . Now we solve 0 = f ( h ) = - πh 2 + (4 / 3) πhR , getting h = 0 or h = 4 R/ 3. We compute V (0) = V (2 R ) = 0 and V (4 R/ 3) = (32 / 81) πR 3 . The maximum is the latter; since the volume of the sphere is (4 / 3) πR 3 , the fraction of the sphere occupied by the cone is (32 / 81) πR 3 (4 / 3) πR 3 = 8 27 30% . EXAMPLE 6.11 You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is N times as expensive (cost per unit area) as the material used for the lateral side of the cylinder. Find (in terms
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Unformatted text preview: of N ) the ratio of height to base radius of the cylinder that minimizes the cost of making the containers. Let us first choose letters to represent various things: h for the height, r for the base radius, V for the volume of the cylinder, and c for the cost per unit area of the lateral side of the cylinder; V and c are constants, h and r are variables. Now we can write the cost of materials: c (2 πrh ) + Nc (2 πr 2 ) . Again we have two variables; the relationship is provided by the fixed volume of the cylinder: V = πr 2 h . We use this relation to eliminate h (we could eliminate r , but it’s a little easier if we eliminate h , which appears in only one place in the above formula for cost). The result is f ( r ) = 2 cπr V πr 2 + 2 Ncπr 2 = 2 cV r + 2 Ncπr 2 ....
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